Diffraction Patterns Due To A ‘Single-Slit’ And A ‘Circular Aperture

Diffraction Patterns Due to a ‘Single-Slit’ and a ‘Circular Aperture - Example 1

Diffraction patterns can be observed when waves pass through a narrow opening or an obstacle.
Two common examples of diffraction patterns are the patterns produced by a single slit and a circular aperture.
In this example, we will study the diffraction pattern produced by a single slit.
Let’s consider a monochromatic light source emitting waves of wavelength λ.
When the waves pass through a single slit of width a, a diffraction pattern is formed on a screen placed at a distance L from the slit.

Conditions for Diffraction

For visible light, the width of the slit should be of the order of the wavelength of light (a ~ λ).
The distance between the screen and the slit should be much larger than the width of the slit (L » a).
The screen should be placed perpendicular to the incident waves.

Intensity Distribution in the Diffraction Pattern

The intensity of the diffracted waves varies with the angle of deviation θ.
The intensity distribution in the diffraction pattern is given by the equation: I = I₀ * (sin(α)/α)², where I₀ is the intensity of the central maximum and α = (πa*sin(θ))/λ.

Single-Slit Diffraction Pattern

The central maximum is the brightest spot in the diffraction pattern and is observed at θ = 0.
The intensity decreases as we move away from the central maximum on both sides.
The first minimum is observed at θ = λ/a, and the next minima occur at multiples of this angle.
The width of the central maximum is larger compared to the minima.

Example: Single-Slit Diffraction

Let’s consider a diffraction pattern formed by a monochromatic light source with a wavelength of 600 nm (visible light).
The single slit has a width of 0.01 mm.
The screen is placed 2 m away from the slit.
Calculate the angle at which the first minimum of the diffraction pattern occurs.

Solution

Given:

Wavelength of light, λ = 600 nm = 600 × 10⁻⁹ m
Width of the single slit, a = 0.01 mm = 0.01 × 10⁻³ m
Distance between slit and screen, L = 2 m We need to find the angle of the first minimum, θ.

Calculation

Using the formula α = (πa*sin(θ))/λ, we can rearrange the equation to find θ: sin(θ) = (α * λ) / (πa) θ = arcsin((α * λ) / (πa))

Substituting Values

Plugging in the values: θ = arcsin(((π * 0.01 × 10⁻³ * 600 × 10⁻⁹)) / (π * 0.01 × 10⁻³))

Calculation (continued)

Simplifying the expression: θ = arcsin((6 × 10⁻⁶) / (0.01 × 10⁻³)) θ = arcsin(6 × 10⁻³) θ = 0.344 degrees

Answer

The angle at which the first minimum of the diffraction pattern occurs is approximately 0.344 degrees.

Conclusion

Diffraction patterns are formed when waves pass through a narrow opening or an obstacle.
The intensity distribution in the diffraction pattern can be calculated using the equation I = I₀ * (sin(α)/α)².
In the case of a single slit, the diffraction pattern consists of a central maximum and alternating minima.
In the example, we calculated the angle at which the first minimum occurs using the given parameters.

Properties of Diffraction Patterns Due to a Single-Slit

The central maximum is the brightest spot in the diffraction pattern.
The intensity of the central maximum drops off as the angle increases.
The width of the central maximum is larger compared to the minima.
The minima are regions of complete destructive interference, resulting in low or zero intensity.
The diffraction pattern exhibits a series of alternating maxima and minima.

Mathematical Analysis of Diffraction Pattern Due to a Single-Slit

The intensity distribution in the diffraction pattern is given by the equation: I = I₀ * (sin(α)/α)².
I is the intensity at a certain angle, α is given by (πa*sin(θ))/λ, and I₀ is the intensity at the central maximum.
The intensity is normalized using the value of I₀.

Factors Affecting Diffraction Pattern Due to a Single-Slit

The width of the slit (a) affects the angular width of the central maximum and the distance between the consecutive minima.
As the width of the slit decreases, the angular width of the central maximum increases, resulting in a narrower diffraction pattern.
The wavelength of the waves (λ) affects the overall size of the diffraction pattern.
A smaller wavelength leads to a smaller diffraction pattern.
The screen-to-slit distance (L) affects the spacing between the minima.
Larger values of L result in larger distances between the minima.

Circular Aperture Diffraction Pattern

The diffraction pattern produced by a circular aperture is similar to that of a single slit diffraction pattern.
The intensity distribution follows a similar pattern, with a central maximum and multiple minima.
However, the angular width of the central maximum and the distances between minima are different compared to a single slit.

Applications of Diffraction Patterns

Diffraction patterns are commonly observed in everyday life.
They are responsible for the spread of light around objects, resulting in phenomena such as shadows and color fringes.
Diffraction patterns are utilized in various scientific fields, such as microscopy, X-ray crystallography, and astronomy.
They also play a significant role in determining the structural characteristics of materials.

Example: Diffraction Pattern Due to a Circular Aperture

Consider a circular aperture with a diameter of 5 mm.
The wavelength of the incident waves is 500 nm.
Calculate the angle at which the first minimum occurs.

Solution

Given:

Diameter of the circular aperture, d = 5 mm = 5 × 10⁻³ m
Wavelength of the waves, λ = 500 nm = 500 × 10⁻⁹ m We need to find the angle of the first minimum, θ.

Calculation

For circular apertures, the angle of the first minimum can be calculated using the equation: θ = 1.22 * λ / d θ = 1.22 * 500 × 10⁻⁹ / (5 × 10⁻³) θ = 0.122 radians

Answer

The angle at which the first minimum of the diffraction pattern occurs for a circular aperture with a diameter of 5 mm and a wavelength of 500 nm is approximately 0.122 radians.

Conclusion

Diffraction patterns due to a single slit and a circular aperture exhibit characteristic intensity distributions.
The central maximum is the brightest spot in the pattern, and the intensity decreases as the angle increases.
The width of the central maximum is larger compared to the minima.
The size and spacing of the diffraction pattern are influenced by factors such as the width of the slit or the aperture, the wavelength, and the distance between the source and the screen.
Applications of diffraction patterns can be found in various scientific fields.

Single-Slit Diffraction Pattern - Huygens Principle

The Huygens Principle states that every point on a wavefront acts as a source of secondary spherical wavelets.
In the case of single-slit diffraction, each point on the slit acts as a source of secondary waves.
The secondary waves interfere constructively and destructively, leading to the observed diffraction pattern.

Example: Diffraction Pattern Due to a Single-Slit

Let’s consider a diffraction pattern formed by a monochromatic light source with a wavelength of 650 nm.
The single slit has a width of 0.02 mm.
The screen is placed 3 m away from the slit.
Calculate the angle at which the first minimum of the diffraction pattern occurs.

Solution

Given:

Wavelength of light, λ = 650 nm = 650 × 10⁻⁹ m
Width of the single slit, a = 0.02 mm = 0.02 × 10⁻³ m
Distance between slit and screen, L = 3 m We need to find the angle of the first minimum, θ.

Calculation

Using the formula α = (πa*sin(θ))/λ and θ = arcsin((α * λ) / (πa)), we can proceed as follows: θ = arcsin(((π * 0.02 × 10⁻³ * 650 × 10⁻⁹)) / (π * 0.02 × 10⁻³)) Simplifying the expression: θ = arcsin((13 × 10⁻⁵) / (0.02 × 10⁻³)) θ = arcsin(0.65) θ = 39.3 degrees

Answer

The angle at which the first minimum of the diffraction pattern occurs is approximately 39.3 degrees.

Circular Aperture Diffraction Pattern - Bessel’s Function

The intensity distribution in the diffraction pattern produced by a circular aperture is described by Bessel’s function.
Bessel’s function represents the interference of the secondary waves emerging from different points on the circular aperture.
The central maximum in the diffraction pattern is surrounded by alternating bright and dark rings.

Example: Diffraction Pattern Due to a Circular Aperture

Consider a circular aperture with a diameter of 2 mm.
The wavelength of the incident waves is 600 nm.
Calculate the angle at which the third dark ring in the diffraction pattern occurs.

Solution

Given:

Diameter of the circular aperture, d = 2 mm = 2 × 10⁻³ m
Wavelength of the waves, λ = 600 nm = 600 × 10⁻⁹ m We need to find the angle at which the third dark ring occurs, θ.

Calculation

For circular apertures, the radius of the dark ring can be calculated using the equation: r = 1.22 * sqrt(n * λ * L) To find θ, we can use the relation: θ = arcsin(r / L) Substituting the values: θ = arcsin((1.22 * sqrt(3 * 600 × 10⁻⁹ * L)) / L) θ = arcsin(1.22 * sqrt(3 * 600 × 10⁻⁹)) θ = arcsin(3.337 × 10⁻⁵) θ ≈ 1.9 degrees

Answer

The angle at which the third dark ring in the diffraction pattern occurs for a circular aperture with a diameter of 2 mm and a wavelength of 600 nm is approximately 1.9 degrees.